Chapter 4, Problem 44A

To determine

The initial acceleration of the wishbone.

Answer to Problem 44A

  $13{\text{m/s}}^2$

Explanation of Solution

Given:

Both K and G pull the wishbone towards themselves (both of them pull in the opposite direction).

The mass of the wishbone is $m=13\text{g}=13\text{g}\times \left(\frac{1kg}{1000\text{g}}\right)=0.013\text{kg}$ .

Formula used:

Newton’s Second Law: The object’s acceleration is equivalent to the sum of the forces acting on the object divided by the object’s mass. That is,

  $\begin{array}{l}a=\frac{F_{\text{net}}}{m}\\ F_{\text{net}}=ma\end{array}$

Calculation:

Consider G pulls the wishbone with a force $F$ towards himself.

Consider K pulls the wishbone with a force $F+0.17\text{N}$ towards himself (opposite direction to G).

Find the net force $\left(F_{\text{net}}\right)$ acting on the wishbone as follows:

  $\begin{array}{c}{F}_{\text{net}}=\left(F+0.17\right)-F\\ =0.17\text{N}\end{array}$

Find the initial acceleration of the wishbone as follows:

Apply Newton’s Second Law,

  $\begin{array}{l}a=\frac{F_{\text{net}}}{m}\\ a=\frac{0.17\text{N}\times \left(\frac{1\text{kg}\cdot {\text{m/s}}^2}{1\text{N}}\right)}{0.013\text{kg}}\\ a=\frac{0.17\text{kg}\cdot {\text{m/s}}^2}{0.013\text{kg}}\\ a=13{\text{m/s}}^2\end{array}$

Conclusion:

Thus, the initial acceleration of the wishbone is $13{\text{m/s}}^2$ .